(2r^2cosθsinθ + rcosθ)dθ + (4r + sinθ - 2rcos^2θ)dr = 0

Name: (2r^2cosθsinθ + rcosθ)dθ + (4r + sinθ - 2rcos^2θ)dr = 0
Uploaded: 2023-04-12T10:59:03-08:00
Duration: 2 min 2 s
Channel: Numerade
Description: (2r^2cosθsinθ + rcosθ)dθ + (4r + sinθ - 2rcos^2θ)dr = 0

Question

Madhur L · Answer

Hi, given R of t is equal to 60i plus 3 t j plus 2 t k. We need to find curvature of this curve.

Adi S · Answer

So, let us start with the concept which we are going to use here for this question. Integration

Sri K · Answer

The following question G, that is columns 1 through 12. So the value is 0 0 .0374 minus 0 .0080,

Adi S · Answer

let let the two cylinders be the two cylinders be f of x y z is equals to x square plus y square

Adi S · Answer

In this problem we are asked to solve the initial value problem 4 times x squared times d square

Question

\((2r^2\cos\theta\sin\theta + r\cos\theta)d\theta + (4r + \sin\theta - 2r\cos^2\theta)dr = 0\)

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